A) \[y+2x=\frac{{{c}^{2}}{{x}^{2}}}{y}\]
B) \[y-2x=\frac{{{c}^{2}}{{x}^{2}}}{y}\]
C) \[y+2x={{c}^{2}}{{x}^{2}}y\]
D) None of these
Correct Answer: C
Solution :
\[{{x}^{2}}dy+y(x+y)dx=0\] \[\Rightarrow \] \[{{x}^{2}}dy=-y(x+y)dx\] \[\Rightarrow \] \[\frac{dy}{dx}+\frac{y(x+y)}{{{x}^{2}}}=0\] Put \[y=vx\Rightarrow \frac{dy}{dx}=v+\frac{dv}{dx}.x\] \[v+\frac{dv}{dx}.x+\frac{vx(x+vx)}{{{x}^{2}}}=0\] \[\Rightarrow \] \[v+x\frac{dv}{dx}+v(1+v)=0\] \[\Rightarrow \] \[x\frac{dv}{dx}=-(2v+{{v}^{2}})\] \[\Rightarrow \] \[\frac{dv}{v(v+2)}=-\frac{dx}{x}\] \[\Rightarrow \] \[\frac{1}{2}\left( \frac{1}{v}-\frac{1}{v+2} \right)dv=-\frac{1}{x}dx\] On integrating, \[\frac{1}{2}[\log v-\log (v+2)]=-\log x-\log c\] \[\Rightarrow \] \[log\text{ }v-log(v+2)+2\text{ }log\text{ }xc=0\] \[\Rightarrow \] \[log\text{ }{{x}^{2}}{{c}^{2}}v=log(v+2)\] \[\Rightarrow \] \[{{x}^{2}}{{c}^{2}}\frac{y}{x}=\frac{y}{x}+2\] \[\Rightarrow \] \[{{c}^{2}}yx=\frac{y+2x}{x}\] \[\Rightarrow \] \[{{x}^{2}}y{{c}^{2}}=y+2x\]
You need to login to perform this action.
You will be redirected in
3 sec
